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What would happen if the force of gravitation suddenly starts varying as $1$$/$$r^3$ instead of $1/r^2$ ? Would the symmetry of universe now seen be disrupted?

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If you're asking physics questions to satisfy your own curiosity at 16 years old, you don't need to apologize for how it sounds. Good luck on your continuing education. –  AdamRedwine Oct 18 '12 at 11:53
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Not a direct answer but might be of interest. There are three cases of a central power law where all the orbits can be written down explicitly using elementary functions---inverse square, direct proportionality (Hooke's law) and the $\frac 1{r^3}$ of your question. In the latter case, the orbits are the so-called Cotes' spirals about which the internet is full of information. By the way, all three were known to Newton and are dealt with in his "Principia". –  mathuser4891 Jun 18 '13 at 9:58
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2 Answers 2

It depends on what you mean by symmetry, But the force law $1/r^3$ does not have stable orbits. Small perturbations will de stabilize your orbits to fall inward or outward.

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As addition to this answer: http://en.wikipedia.org/wiki/Bertrand%27s_theorem. –  NikolajK Oct 18 '12 at 11:45
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Also the iron sphere theorem would not hold as it relies on area of a sphere increasing with $r^2$. Relativity gets around this by changing the geometry. –  SMeznaric Oct 18 '12 at 15:54
    
@NickKidman Wow! this and Bertrand's theorem are amazing! Thanks for the links. See also Dimension10's point 2 below and the Wiki page linked therein. –  WetSavannaAnimal aka Rod Vance Aug 29 '13 at 5:04
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Such a change can only occur if space becomes 4 dimensional instead of 3 dimensional (see for example, my answer here). That will have quite profound implications. For example,

  1. The Riemann Curvature Tensor will have one more part (but this "part" means many more components) along with the Ricci Curvature Tensor and the Weyl Curvature Tensor. This actually related to the Electromagnetic force so in toher words, Gravity and EM will behave in the same way.

  2. The 4+1=5 dimensional spacetime is actually unstable (agreedly, I don't know why, but whatever), at least according to Wikipedia's "privileged character of 3+1 dimensional spacetime" article.

  3. The string theory landscape maybe a bit smaller (I think.).

  4. The Ricci curvature in a vacuum on an Einstein Manifold would no longer be exactly $\Lambda g_{ab}$. There will be an ugly coefficient of $\frac{2}{3} $

  5. The magnetic field could not be written as a vector, unlike the electric field. This is because it would have 6 components whereas the spatial dimension is only 4. So, perhaps humans would be more familiar with exterior algebras, earlier than us who live in 3+1=4 dimensions. Either that or we would be trying to find out how magnetism works. Or we would just die out, due to Implication (2).

  6. Cross Products would be harder to evaluate. Another reason why we would be more familiar with exterior algebras!

  7. Ok, so I'll come back to your question now:

So in 4D space the Newtonian gravitational force is: $$\vec F=G_5 \frac{m_1m_2}{r^3}$$ Here, $G_5$ is the gravitational constant in 4 spatial dimensions (5 spacetime dimensions, thus the 5 subscript.). It has to be different from the ordinary gravitational constant because it needs different units (since the $\frac{1}{r^3}$ immediately induces an extra units of $\operatorname{m}^{-1}$, so the Gravitational constant needs to be adjusted to have a units with extra $\operatorname{m}$, i.e $\frac{\operatorname{m}^4}{\operatorname{kg} \operatorname{s^2}}$. According to string theory, the gravitational constant gets multiplied by the string length for every extra dimension added: $$G_5=\ell_s G_4$$

The exact string length is obviously unknown (a free parameter of string theory, as far as I know!) but let us just say it is $1.2374713757$ multiplied by the 3+1=4-dimensional (i.e ordinary) planck length (it would be a coincidence if it were exact!). The 3+1=4-dimensional planck length is approximately $1.616199\times 10^{-35}$. Then, for the estimate, the string length would be approximately $2\times 10^{-35}$ (now you know why I chose 1.2374713757 !:) (not a factorial sign )). Then, $G_5\approx 2\times 10^{-35}\times6.673\times 10^{-11}=13.346\times 10^{-46}$

That is very weak gravity! Now, how heavy is an apple? It has a mass of say $0.1\operatorname{kg}$. Then, the gravitational force between that and the earth (in 5-dimensional spacetime,/4-dimensional space), is approximately around (after applying both the inverse cube proportion and the reduced, very weak, small, tiny, gravitational constant.),:: $$F\approx 4.82\times10^{-32} \operatorname{Newtons}$$$$ And in daily life, it is like 1 Newton!..

Finally, in summary, gravity, would be much weaker!.

P.S. Your question is actually very relevant because recently, the inverse-square law has been tested by some experiments http://arxiv.org/abs/hep-ph/0011014.

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Minor comment to the answer (v2): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/hep-ph/0011014 –  Qmechanic Jun 17 '13 at 11:42
    
@Qmechanic: Thanks for telling me. I edited my answer accordingly. –  Dimensio1n0 Jun 17 '13 at 11:47
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